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User blog:Mh314159/Rank and Argument Strings
Please comment on my latest attempt to climb toward TREE(3) as a personal intellectual challenge. Thank you for your kind attention! Conceptually, here's how this works. There is a base function, which I chose to be diagonalized Ackermann numbers, but it probably doesn't have to be that strong once the line gets long. Then there is a line of bracket-enclosed numbers or comma-separated strings of numbers. The last two brackets expressions and the "rank" and "argument". All others to their left are "orders". There are rules that recurse longer and more complex expressions down to simpler expressions. In order from most complex to least complex, expression recurse in the following order of rule sets: o -> ss -> sn -> ns -> nn -> base. I think this generates really big numbers, much bigger than anything else I have tried to do (although I actually find my vertical bar notation easier and more fun to play with). I am not done with this because I a planning another rule set to diagonalize the entire notation and generate extremely long rows of orders. Here are the rule sets with some examples. For two sets of brackets, the number or string in the left brackets is the "rank" of the function and the number or string in the right brackets is the "argument" Rule set nn (both bracket sets contain numbers_ for a = 1, n1 = n+1 for a > 1, 1a = a↑a a na = n-1na-1a functional recursion notation 22 = 1212 = 132 = 1112 = 114 = 1(4↑44] (This is approximately G2 (slightly greater) in the Graham's number sequence) 23 = 12(2]3 = 114↑↑↑↑43 (This is greater than the G2 th iteration of the Graham's number sequence = G(G2) ) 24 = 12(3](4] > G(G(G2)) 32 = 23(1]2 = 242 = 22222 = 222N with N = 22 = 14↑44 as above and where 2N recurses as above into a 1-functional recursion pyramid N layers high; i.e., find 21 = 3 so find 132. Next find 1P3 where P = 132 and repeat this process N times. Then to find 32 apply the 2 function two more times. Rule set ns bracket contains number, right bracket contains string) [na,b,c,...,p,0 = na,b,c,...,p drop trailing zeroes na,b = [na,b-1]]a,b-1 na,b,c,...p = [n[a,na,b,na,b,c,...,p-1]]a,b,c,...p-1 12,1 = [12]2 = 42. 13,2 = [13,1]3,1 = [[13,1]3]3. 13,4,5,6 = [1[3,13,4,13,4,5,5]]3,4,5,5 and if 1[3,13,4,1 3,4,5,5] = n, the next step is [n[3,n3,4,n3,4,5,4]]3,4,5,4 etc. Rule set sn (left brack contains string, right bracket contains number) n,m,...,p,ra = na,a,...,a #a = [na,n,ma,...,n,m,...pa,r-1]a 1,12 = 12,2,...,2 instances of 2 = [12]2 = 42 2,14 = 24,4,...,4 instances of 4 = [24]4 2,24 = 24,4,...,4 instances of 4 = [24,1]4 which = [24]4,4,...,4 where instances of 4 = [[24]4]4 2,3,45 = 25,5,...5 #5 = [25,2,35,3]5 Rule set ss (left brack contains string, right bracket contains string) Define x = string a,b,...p Define y = string q,r,...v Define y' = string q,r,...v-1 Define z = string y' with each term except the last replaced by xy' drop trailing zeroes xy = x,x,...xz with xy' instances of x. Eventually, z will recurse to an expression of the form stringnumber and then that rule set is used to generate an expression of the form numberstring and eventually a number is the result, and then we know the number of x's in x,x,...xz and the sn and ns rule sets then apply Example where x always refers to the string in the immediately preceding xy expression: 2,34,3 = 2,3,2,3...[2,34,2,2] with instances of 2,3 = 2,34,2 2,34,2 = 2,3,2,3...[2,34,1,1] with instance of 2,3 = 2,34,1 2,34,1 = 2,3,2,3...[2,34] with instances of 2,3 = 2,34 and then the sn rule set applies and if 2,34,1 = n then the previous step can be recursed as 2,34,2 = 2,3,2,3...n,1 with instances of 2,3 = n 2,3,2,3...n,1 = 2,3,2,3...[xn] with instances of 2,3,2,3,... (n instances) in the second expression = xn and the sn rule set determines the value of xn and then the value of 2,34,2, which we will call m. 2,34,3 = 2,3,2,3...m,2 with instances of x = m. And 2,3,2,3...m,2 = 2,3,2,3...[xm,1,1] with instances of x in the second expression = xm,1, etc. Rule set o ab...c where there are three or more sets of brackets and a, b, ... c can be numbers or strings ab...c = b...c if a = 1 if a > 1, then a' = a with its final entry decremented by 1 and ab...c = a'n,n,...n...n,n,...n with n = a'b...c and with n instances of n in each bracket 222 = n,n,...nn,n,...n with n = 22 repeated n times in each bracket (note that 122 = 22 and that 1n,n,...nn,n,...n = n,n,...nn,n,...n) 322 = 2n,n,...nn,n,...n with n = 222 repeated n times in each bracket and where 2n,n,...nn,n,...n = 1m,m,...mm,m,...m where m = 1n,n,...nn,n,...n and is repeated m times 2,222 = 2,1n,n,...nn,n,...n with n = 2,122 repeated n times in each bracket and where 2,1n,n,...nn,n,...n = 2m,m,...mm,m,...m where m = 2n,n,...nn,n,...n and is repeated m times Category:Blog posts